Problem: The local linear approximation to the function $h$ at $x=-4$ is $y=-4x-1$. What is the value of $h(-4)+h'(-4)$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $10$ (Choice B) B $11$ (Choice C) C $12$ (Choice D) D $13$
Solution: The local linear approximation of $h$ at $x=-4$ is achieved using the equation of the line tangent to $h$ at $x=-4$. In other words, $y=-4x-1$ is the equation of the line tangent to the graph of $h$ at $x=-4$. How can we use this to find $h(-4)$ and $h'(-4)$ ? Since the line is tangent to the graph of $h$ at $x=-4$, we know two key facts about it: The line passes through the point $({-4},{h(-4)})$ The line's slope is ${h'(-4)}$ The slope of $y={-4}x-1$ is ${-4}$. The $y$ -value that corresponds to $x={-4}$ is $-4({-4})-1={15}$. Now we can find our answer: ${h(-4)}+{h'(-4)}={15}+({-4})=11$